package cn.lccabc.dichotomy.no0033;

/**
 * 假设按照升序排序的数组在预先未知的某个点上进行了旋转。
 *
 * ( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。
 *
 * 搜索一个给定的目标值，如果数组中存在这个目标值，则返回它的索引，否则返回 -1 。
 *
 * 你可以假设数组中不存在重复的元素。
 *
 * 你的算法时间复杂度必须是 O(log n) 级别。
 *
 * 示例 1:
 *
 * 输入: nums = [4,5,6,7,0,1,2], target = 0
 * 输出: 4
 * 示例 2:
 *
 * 输入: nums = [4,5,6,7,0,1,2], target = 3
 * 输出: -1
 *
 */
public class Solution {

    public static int search(int[] nums, int target) {
        if (nums.length == 0) return -1;
        int left = 0, right = nums.length - 1;
        if (nums[nums.length - 1] < nums[0]){
            int l = 0, r = nums.length - 1;
            while (l + 1 < r){
                int mid = l + (r - l) / 2;
                if (nums[mid] > nums[l]){
                    l = mid;
                } else {
                    r = mid;
                }
            }
            if (target >= nums[0]){
                right = l;
            } else {
                left = r;
            }
        }
        while (left <= right){
            int mid = left + (right - left) / 2;
            if (nums[mid] == target){
                return mid;
            } else if (nums[mid] < target){
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return -1;
    }


    public static void main(String[] args) {
        int[] nums = new int[]{4,5,6,7,0,1,2};
        int target = 3;
        int result = search(nums, target);
        System.out.println(result);
    }
}
